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Choice Is Hard

机译:选择很难

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Let P = {C_1, C_2,...,C_n} be a set of color classes, where each color class C_i consists of a pair of objects. We focus on two problems in which the objects are points on the line. In the first problem (rainbow minmax gap), given P, one needs to select exactly one point from each color class, such that the maximum distance between a pair of consecutive selected points is minimized. This problem was studied by Consuegra and Narasimhan, who left the question of its complexity unresolved. We prove that it is NP-hard. For our proof we obtain the following auxiliary result. A 3-SAT formula is an LSAT formula if each clause (viewed as a set of literals) intersects at most one other clause, and, moreover, if two clauses intersect, then they have exactly one literal in common. We prove that the problem of deciding whether an LSAT formula is satisfiable or not is NP-complete. We present two additional applications of the LSAT result, namely, to rainbow piercing and rainbow covering. In the second problem (covering color classes with intervals), given P, one needs to find a minimum-cardinality set I of intervals, such that exactly one point from each color class is covered by an interval in I. Motivated by a problem in storage systems, this problem has received significant attention. Here, we settle the complexity question by proving that it is NP-hard.
机译:让P = {C_1,C_2,...,C_N}是一组颜色类,其中每个颜色类C_I由一对对象组成。我们专注于两个问题,其中对象在线上的点。在第一个问题(彩虹Minmax间隙)中,给定P,需要从每个颜色类别选择一个点,使得一对连续选择点之间的最大距离最小化。 CONSUEGRA和NARASIMHAN研究了这个问题,他尚未解决了其复杂性的问题。我们证明它是NP-HARD。对于我们证明,我们获得以下辅助结果。如果每个条款(被视为一组文字)在大多数其他条款中相交,那么,一个3-SAT公式是一个LSAT公式,而且,如果两个条款相交,那么他们就是一个共同的一个文字。我们证明了决定LSAT公式是否满足或不完全的问题。我们展示了LSAT结果的两个额外应用,即彩虹刺穿和彩虹覆盖物。在第二个问题(覆盖间隔的颜色类)中,给定P,需要找到最小基数集合I的间隔,从而通过I中的一个间隔涵盖了来自每个颜色类的一个点。通过问题的激励激励存储系统,这个问题受到了重大关注。在这里,我们通过证明它是NP-HARD来解决复杂性问题。

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